# Math Help - Compex number question

1. ## Compex number question

having trouble with this question:

Given that p and q are real and that 1 + 2i is a root of this equation:

z^2 + (p+5i)*z +q(2 - i) = 0

- Determine the values of p and q
- Determine the other root of the equation

I really have no idea how to approach this.

2. Originally Posted by dojo
having trouble with this question:

Given that p and q are real and that 1 + 2i is a root of this equation:

z^2 + (p+5i)*z +q(2 - i) = 0

- Determine the values of p and q
- Determine the other root of the equation

I really have no idea how to approach this.
The other root will be the complex conjugate, you will need this to find $p$ and $q$

3. Originally Posted by dojo
Given that p and q are real and that 1 + 2i is a root of this equation: z^2 + (p+5i)*z +q(2 - i) = 0
- Determine the values of p and q
- Determine the other root of the equation
Originally Posted by pickslides
The other root will be the complex conjugate, you will need this to find $p$ and $q$
No pickslides that is not true. It is true only if the coefficients are real.
In this case they are not real.

@ dojo
Expand $(1+2i)^2+(p+5i)(1+2i)+q(2-i)$ to get $(p+q-13)+i(2p-2q+9)$.
Now you have two equations in two unknowns.

4. Hi plato - should that expansion not be (p + 2q - 13) + i(2p -q +9).

I'm still at a loss - I thought it would be solved by comparing the original equation to (x - α) (x - β). After expanding with replacing with the root and its conjugate fianlly compare the coefficiants.

I then get p = -2 -5i and q = 2 + i

5. Originally Posted by dojo
Hi plato - should that expansion not be (p + 2q - 13) + i(2p -q +9).

I'm still at a loss - I thought it would be solved by comparing the original equation to (x - α) (x - β). After expanding with replacing with the root and its conjugate fianlly compare the coefficiants.

I then get p = -2 -5i and q = 2 + i
Your expansion is correct. But since you're told that p and q are real it is unlikely that your answers are correct ....

Did you equate the real and imaginary parts of the expanded expression to zero and solve the resulting two equations simultaneously (as advised by Plato)? If so, please show your work.

6. I'm lost Mr Fantastic. can you help me further as I dont know what you mean. Here are my workings...

D

7. Originally Posted by dojo
having trouble with this question:

Given that p and q are real and that 1 + 2i is a root of this equation:

z^2 + (p+5i)*z +q(2 - i) = 0

- Determine the values of p and q
- Determine the other root of the equation

I really have no idea how to approach this.

$p + 2q - 13 + i(2p -q +9) = 0 + 0i$.

Equating real and imaginary parts gives:

$p + 2q - 13 = 0$ and $2p - q + 9 = 0$.

Solve these equations simultaneously for $p$ and $q$.

8. Originally Posted by dojo
I'm lost Mr Fantastic. can you help me further as I dont know what you mean. Here are my workings...

D

Have another look at Plato's post

Originally Posted by Plato
No pickslides that is not true. It is true only if the coefficients are real.
In this case they are not real.
The two roots are not conjugate since the coefficients (p+5i and q(2-i)) are not real

Originally Posted by Plato
Expand $(1+2i)^2+(p+5i)(1+2i)+q(2-i)$ to get $(p+q-13)+i(2p-2q+9)$.
Now you have two equations in two unknowns.

9. OK... When I expand I get:

(p + 2q + 15) + i(2p + 9 - q) = 0
(p + 2q -13) + i(9 + 2p -q) = 0

not sure how i would solve using simultaneous eq. though

10. Expanding $(1+2i)^2+(p+5i)(1+2i)+q(2-i)=0$ gives $(p+2q-13)+i(2p-q+9)=0$

Which gives two equations
$p+2q-13=0$ (real part)
$2p-q+9=0$ (imaginary part)

11. Ah-ha I see. so I get p = -1 and q = 7.?

What about finding the other root?

12. You have to know that the roots of a quadratic az²+bz+c=0 are linked by

$z_1 + z_2 = -\frac{b}{a}$

$z_1 \times z_2 = \frac{c}{a}$

You can use either of these relations (the first one is the simplest one)